Adjoints Preserve Limits

A right adjoint cannot lose a limit — which turns "does this functor have an adjoint?" into a check you can finish on one line.

The idea

Theorem (Right adjoints preserve limits).

If $F \dashv G$, then $G$ carries limit cones to limit cones, and $F$ carries colimit cones to colimit cones.

The theorem explains an asymmetry. The underlying set of a direct product of groups is the product of the underlying sets: an element of $G \times H$ is a pair, and nothing else. But the free group on a disjoint union of two sets is not the direct product of the two free groups; it is their free product. One construction survives forgetting and the other does not, because the free functor and the forgetful functor sit on opposite sides of a single bijection. Forgetting is a right adjoint, so it keeps limits — products, equalizers, pullbacks. The free construction is a left adjoint, so it keeps colimits, and nothing obliges it to keep anything else.

Read backwards, the theorem is an obstruction, and that is its main use. A functor that destroys some limit cannot be a right adjoint, so it has no left adjoint. One failed check on a small diagram settles a question that would otherwise demand building the adjoint by hand.

Ways to work on it

• Walkthrough. Why right adjoints preserve limits and left adjoints preserve colimits — and how one failed limit rules out an adjoint.
• Proof. Prove that right adjoints preserve limits by transposing cones across the adjunction.
• Practice. Decide whether a functor preserves a named construction, say which adjoint a failure rules out, and compute a limit of groups on underlying sets.
• Hardest. Rule out adjoints for the power-set functor, catch a functor destroying a coproduct, and pick the exponent law the theorem licenses.

Not sure where to start? Take the ten-question placement test.